# Proof on the inequality involving matrix splitting and trace operator

Suppose positive definite matrices $V, B, D\in\mathbb{R}^{n\times n}$ are given, where $D$ only contains diagonal entries of $V$, i.e., $D=diag(V)$, and $X, G\in\mathbb{R}^{n\times 2}$. Could the following be proved:

$$\frac{1}{2}tr((D^{-1}G)^TVD^{-1}G) - 2tr(X^TV(D^{-1}G)) + 2tr((D^{-1}G)^TBX)<0$$

for some arbitrary $G$, or for the case $G=2(VX-BX)$. I tried to apply the steps from: Possible proof for the relation involving matrix trace

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@user50601. See Wikipedia: Shur complemet (en.wikipedia.org/wiki/Schur_complement) and Matrix Block inversion of positive definite matrices in Jean Gallier Home page (cis.upenn.edu/~jean/schur-comp.pdf ) – MathOverview Feb 29 '12 at 13:54
Surely this inequality is obtained starting from the trace of a 2x2 or 3x3 matrix whose entries are a function of blocks $D,G,V,X,B$. I hope this helps. – MathOverview Feb 29 '12 at 13:57
@Elias Thanks for the answer; I'm impressed by your observation. I've downloaded the paper, but would kindly ask you to be more specific. I'm unable to find the way to establish the inequality. – user506901 Feb 29 '12 at 14:25
@Elias Could the above inequality be somehow cast to $$\frac{1}{2}(D^{-1}g)^TVD^{-1}g - 2x^TV(D^{-1}g) + 2(D^{-1}g)^TBx<0$$for $g, x\in\mathbb{R}^n$, later with the proof extension to the original inequality with trace involved? If so, I guess the proof could be facilitated (?) – user506901 Feb 29 '12 at 16:33

Put $U = [U_1, U_2, U_3] ^ T$ and $$L=\begin {pmatrix} V & 0 & 0 \\ 0 & V & 0 \\ 0 & 0 & B \end {pmatrix}$$ By hipothesis we heve $Tr(U^ TLU) \geq 0.$ The difficulty is now set $U_1$, $U_2$ and $U_3$ properly so that $$-Tr(U^TLU) =\frac{1}{2}tr((D^{-1}G)^TVD^{-1}G) - 2tr(X^TV(D^{-1}G)) + 2tr((D^{-1}G)^TBX)$$ Now you can use knowledge of quadratic forms to get your proof. That is a lot sitema in which the variables are the building blocks$U_1$, $U_2$ solve (he may have more than one solution) for $U_1$, $U_2$ in terms of $[D^{-1} g]$ and $x$.
@user506901. In my answer is easy to see that $U_1=D^{-1}g$. Good Look. – MathOverview Mar 1 '12 at 11:12
I appreciate your help, but it's not easy for me to understand your approach. If you have some time, perhaps you could explain your $U=[U_1, U_2, U_3]$ construction (or, simply, give the solution). – user506901 Mar 1 '12 at 12:18